These notes are still being edited: the text may contain typos, inaccuracies, and places that have not been fully worked out locally. If you find something, please let the author know (contact info on the course page).
Compute:
\((1+2 i)(2-i)+(1-2 i)(2+i)\);
\(\left(\frac{1-i}{1+i}\right)^{3}\).
\((1+2 i)(2-i)+(1-2 i)(2+i) = 2 + 4i -i + 2 + 2 -4i + i + 2 = 8\);
We have
\[
\frac{1 - i}{1 + i} = \frac{(1-i)^2}{(1+i) (1-i)} = \frac{-2i}{2} = -i
\]
I.e. \(\left(\frac{1 - i}{1 + i}\right)^{3} = (-i)^3 = -i^3 = i\).
Compute: \(\frac{5}{1+2 i}+\frac{5}{2-i}\).
Find all roots of the equation:
- \(z+|z+1|+i = 0\);
Let \(z = x + iy\). \(-i - z = \left|z + 1\right| \in [0, +\infty )\), i.e. \(\operatorname {Im}(-i-z) = - 1 - y = 0\), \(y = -1\).
Further, \(\left|z + 1\right| = \left|x + 1 - i\right| = \sqrt{(x+1)^2 + 1}\); \(-i-z = -x\). We obtain
\[
-x = \sqrt{(x+1)^2 + 1} \iff \begin{cases} x \leq 0 \\ x^2 = (x+1)^2 + 1 \end{cases} \iff \begin{cases} x \leq 0 \\ 0 = 2x + 2 \end{cases} \iff x = -1
\]
Find all roots of the equation: \(|z|^{2}-2 i z+2 i = 0\).
Solve the system of equations:
- \(\left\{ \begin{array}{l}|z-2 i| = \left|z\right|, \\ |z-i| = |z-1| ;\end{array}\right.\)
The point \(z = x + iy\) is equidistant from \(0\) and from \(2i\), i.e. it lies on the line \(\operatorname {Im}z = y = 1\). Further, the point \(z\) is equidistant from \(i\) and from \(1\), i.e. it lies on the perpendicular bisector of the segment joining them, on the line \(x = y\).
Solve the system of equations: \(\left\{ \begin{array}{l}\left|z^{2}-2 i\right| = 4, \\ |z+1+i| = |z-1-i| .\end{array}\right.\)
Solve the equation: \(z^{7} = -2 + i\);
We have \(-2 + i = r e^{i \varphi }\), where \(r = \sqrt{(-2)^2 + 1^2} = \sqrt{5}\), \(\varphi = \arctan \left(-\frac{1}{2}\right) + \pi\). Then \[
z^7 = \sqrt{5}\, e^{i\varphi } \iff z = \left(\sqrt{5}\right)^{1/7} \cdot \exp \left(i\frac{\varphi + 2\pi k}{7}\right), \qquad k \in \left\{ 0, 1, \ldots , 6\right\}
\] β all \(7\) values of the root.
7 roots: \(z = \left(\sqrt{5}\right)^{1/7} \cdot \exp \left(i\frac{1}{7}\left(\varphi + 2\pi k\right)\right)\), where \(\varphi = \arctan \left(-\frac{1}{2}\right) + \pi\), \(k \in \left\{ 0,1\ldots ,6\right\}\).
Solve the equation: \(z^{8} = 1+i\).
Determine what curve in the plane is defined by the equation:
\(\operatorname {Re} \frac{1}{z} = \frac{1}{a}(a > 0)\);
\(\operatorname {Im} \frac{z-1}{z+1} = 0\).
Recall that \(z\cdot \overline{z} = \left|z\right|^2\), i.e. \(\frac{1}{z} = \frac{1}{\left|z\right|^2}\overline{z}\),
\[
\operatorname {Re}\frac{1}{z} = \frac{1}{\left|z\right|^2} \operatorname {Re}\overline{z} = \frac{1}{\left|z\right|^2}\operatorname {Re}z = \frac{x}{x^2 + y^2}
\]
We have
\[
\frac{1}{a} = \frac{x}{x^2 + y^2} \iff x^2 - ax + y^2 = 0 \iff \left(x - \frac{a}{2}\right)^2 + y^2 = \left(\frac{a}{2}\right)^2
\]
A circle centered at \(\left(\frac{a}{2}, 0\right)\) with radius \(\frac{a}{2}\) (excluding the point \(z = 0\), where \(\frac{1}{z}\) is undefined).
We have
\[
\frac{z-1}{z+1} = (z-1) \cdot \frac{1}{\left|z+1\right|^2}\overline{(z + 1)} = \frac{(z-1)(\overline{z} + 1)}{\left|z+1\right|^2} = \frac{\left|z\right|^2 + z - \overline{z} - 1}{\left|z+1\right|^2}
\]
I.e.
\[
\operatorname {Im}\frac{z-1}{z+1} = \frac{1}{\left|z+1\right|^2} \operatorname {Im}\left(\left|z\right|^2 + z - \overline{z} - 1\right) = \frac{1}{\left|z+1\right|^2} \operatorname {Im}\left(z - \overline{z}\right) = \frac{1}{\left|z+1\right|^2} 2\operatorname {Im}z
\]
Thus, \(\operatorname {Im} \frac{z-1}{z+1} = 0\) \(\iff\) \(\operatorname {Im}z = 0\) (and \(z \neq -1\), where the expression is undefined).
The circle constructed on the segment \([0, a]\) as diameter (excluding the point \(z = 0\))
The real axis (excluding the point \(z = -1\)).
Determine what curve in the plane is defined by the equation: \(\operatorname {Re} \frac{z-1}{z+1} = 0\).
Determine what curve in the plane is defined by the equation: \(\operatorname {Re} \frac{z-a}{z+a} = 0 \quad (a > 0)\).
Determine what set of points \(z\) in the complex plane satisfies the inequality:
\(0 < \arg \frac{i-z}{z+i} < \frac{\pi }{2}\);
\(|z| > 1-\operatorname {Re} z\);
Let us transform:
\[
\frac{i-z}{z+i} = \frac{(i-z)\overline{(z+i)}}{\left|z+i\right|^2} = \frac{(i-z)(\overline{z}-i)}{\left|z+i\right|^2} = \frac{i\overline{z} + 1 - z\overline{z} + iz}{\left|z+i\right|^2} = \frac{1 - \left|z\right|^2 + 2ix}{\left|z+i\right|^2}
\]
The condition \(0 < \arg w < \frac{\pi }{2}\) means that \(w\) lies in the open first quadrant, i.e. \(\operatorname {Re}w > 0\) and \(\operatorname {Im}w > 0\):
\[
\begin{cases} 1 - \left|z\right|^2 > 0 \\ 2x > 0 \end{cases} \iff \begin{cases} \left|z\right| < 1 \\ \operatorname {Re}z > 0 \end{cases}
\]
β the right half of the unit disk, boundary excluded.
Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).
Plan: TikZ: the right half of the unit disk (the set 0 < arg((i-z)/(z+i)) < pi/2), boundaries dashed. Interactive: two sliders for the bounds of the argument sector (currently 0 and pi/2) that recompute the shaded region; on hovering over a point z, show w = (i-z)/(z+i) on a second panel along with its argument.
Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/110/010_arg_halfdisk_fig.tex
If \(\operatorname {Re}z = x > 1\), then the right-hand side \(1 - x\) is negative and the inequality holds. For \(x \leq 1\) both sides are nonnegative, so let us square:
\[
\left|z\right| > 1 - x \iff x^2 + y^2 > 1 - 2x + x^2 \iff y^2 > 1 - 2x
\]
(for \(x > 1\) the inequality \(y^2 > 1 - 2x\) also holds, since the right-hand side is negative). In total: the set \(\left\{ y^2 > 1 - 2x\right\}\) is the part of the plane on the same side of the parabola \(y^2 = 1 - 2x\) as the point \(z = 1\); the parabola itself is excluded.
Figure in development. Will be recreated as an interactive figure (Plotly/Three.js).
Plan: TikZ: the region |z| > 1 - Re z (to the right of the parabola y^2 = 1-2x), parabola dashed, point z = 1. Interactive: a slider for the parameter c in the inequality |z| > c - Re z (the family of parabolas y^2 = c^2 - 2cx), with a check of whether a draggable point belongs to the region.
Source: courses/nedosh_courses/courses/01-analysis/CA/contents/specials/110/020_parabola_region_fig.tex
The right half of the disk of radius 1 centered at the point \(z = 0\).
The part of the plane lying on the same side of the parabola \(y^{2} = 1-2 x\) as the point \(z = 1\) (and bounded by this parabola).
In all cases the points of the boundary curves are excluded.
Determine what set of points \(z\) in the complex plane satisfies the inequality: \(\operatorname {Re} z^{4} > \operatorname {Im} z^{4}\).
Let \(A\) and \(C\) be real, and \(B\) a complex constant, and suppose \(A C < |B|^{2}\). Prove that the equation \[
A|z|^{2}+\overline{B} z+B \overline{z}+C = 0 \quad (A > 0)
\] is the equation of a circle, and find the center of this circle and its radius.
Prove that three pairwise distinct points \(z_{1}, z_{2}, z_{3}\) lie on the same line if and only if the quantity \(\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\) is real.
Prove convergence of the sequence and find its limit:
\(\left\{ \frac{a^{n}}{1+a^{2 n}}\right\} ,|a| < 1\);
\(\left\{ \frac{a^{n}}{1+a^{2 n}}\right\} ,|a| > 1\);
\(\left\{ \frac{1}{n}\left(1+e^{i \varphi }+\ldots +e^{i n \varphi }\right)\right\} , 0 < \varphi < 2 \pi\);
Prove convergence of the sequence and find its limit: \[
\left\{ \frac{1}{\sqrt{n}}\left(1-e^{i \varphi }+e^{2 i \varphi }-\ldots +(-1)^{n} e^{i n \varphi }\right)\right\} ,-\pi < \varphi < \pi
\] .
Let \(\lim_{n \rightarrow \infty } z_{n} = A \neq \infty\). Prove that \[
\lim _{n \rightarrow \infty } \frac{z_{1}+z_{2}+\ldots +z_{n}}{n} = A .
\]
Find the real and imaginary parts of the complex numbers:
\(\left(\frac{1-i}{1+i}\right)^{2025}\);
\((1+i)^{n}-(1-i)^{n}, n \in \mathbb {N}\).
Solve the equations:
\(z^{2}-3 z+(3+i) = 0\);
\(z^{6}-4 z^{3}+8 = 0\).
On the unit circle \(|z| = 1\) two points \(z_1\) and \(z_2\) are taken, \(z_1+z_2 \neq 0\), and tangent lines to the circle are drawn through them. Find the point where these tangent lines intersect.
Regarding the numbers \(z_1, z_2 \in \mathbb {C} \backslash \left\{ 0\right\}\) as vectors in the complex plane, show that \(\operatorname {Re}\left(\overline{z_1} z_2\right)\) is their dot product, and \(\left|\operatorname {Im}\left(\overline{z_1} z_2\right)\right|\) is the area of the parallelogram with sides \(z_1\) and \(z_2\).
Let \(\omega_{0}, \omega_{1}, \omega_{2}, \ldots , \omega_{n-1}\) be the \(n\)-th roots of \(1\), \(n \geq 2\). Prove that the following relations hold \[
\sum _{k = 0}^{n-1} \omega _{k} = 0, \quad \prod _{k = 0}^{n-1} \omega _{k} = (-1)^{n-1}
\]
Prove that the points \(z_{1}, z_{2}, z_{3}\) are the vertices of an equilateral triangle if and only if the condition \(z_{1}^{2}+z_{2}^{2}+z_{3}^{2} = z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\) holds.
These notes are still being edited: the text may contain typos, inaccuracies, and places that have not been fully worked out locally. If you find something, please let the author know (contact info on the course page).
A function \(f: U(z_0) \to \mathbb {C}\) is called differentiable at the point \(z_0 \in \mathbb {C}\) if there exist a number \(A \in \mathbb {C}\) and a function \(\alpha (\Delta z) \xrightarrow [\Delta z \to 0]{} 0\) (\(\Delta z := z - z_0\)), such that
\[
\Delta f = f(z) - f(z_0) = A \cdot \Delta z + \alpha (\Delta z) \cdot \Delta z, \qquad z \in U(z_0)
\]
\(A = \lim_{\Delta z \to 0} \frac{\Delta f}{\Delta z} =: f'(z_0)\).
\(f(z) = u(z) + iv(z) = u(x,y) + iv(x,y)\), where \(u,v: \mathbb {R}^{2} \to \mathbb {R}\).
Theorem 1 (The Cauchy-Riemann equations) Let \(f: U(z_0) \to \mathbb {C}\), \(z_0 = x_0 + iy_0 \in \mathbb {C}\). \(f\) is differentiable at \(z_0\) if and only if
\(u\), \(v\) are differentiable at \((x_0, y_0)\) (in the real sense);
\(u_x = v_y\), \(u_y = -v_x\).
Find all points \(z = x+i y\) at which the following functions are differentiable:
\(x^{2}-i y^{2}\);
\(z \operatorname {Re} z\).
\(u(x,y) = x^2\), \(v(x,y) = -y^2\). \(u_x = 2x\), \(v_y = -2y\). \(u_y = 0\), \(v_x = 0\).
\(f(z) = (x + iy) \cdot x = x^2 + ixy\), \(u(x,y) = \operatorname {Re}f(z) = x^2\), \(v(x,y) = \operatorname {Im}f(z) = xy\). They are differentiable everywhere. The Cauchy-Riemann conditions:
\[
\begin{cases} \frac{\partial u}{\partial x} = 2x = x = \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} = 0 = -y = -\frac{\partial v}{\partial x} \end{cases}
\]
I.e. differentiable only when \(x = y=0\).
on the line \(\operatorname {Re} z = -\operatorname {Im} z\);
Geometric meaning of the Cauchy-Riemann conditions:
\[
\begin{align} u_x + iv_x & = \lim _{\Delta x \to 0} \frac{f(x_0 + \Delta x + iy_0) - f(x_0 + iy_0)}{\Delta x} = \lim _{i\Delta y \to 0} \frac{f(x_0 + i(y_0 + \Delta y)) - f(x_0 + iy_0)}{i\Delta y} = \\ & = \lim _{\Delta y \to 0} \frac{\Delta u + i\Delta v}{i\Delta y} = \lim _{\Delta y \to 0} \frac{\Delta v - i\Delta u}{\Delta y} = v_y - iu_y \end{align}
\]
The partial derivatives are equal only along 2 directions. And this is sufficient for differentiability.
Another interpretation of the Cauchy-Riemann conditions: introduce new βcoordinatesβ: \((x,y) \mapsto (z, \overline{z})\),
\[
\begin{cases} z = x + iy \\ \overline{z} = x - iy \end{cases}, \qquad \begin{cases} x = \frac{z + \overline{z}}{2} \\ y = \frac{z - \overline{z}}{2i} \end{cases}
\]
We have
\[
df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = \frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \overline{z}} d\overline{z}
\]
What are \(\frac{\partial f}{\partial z}\), \(\frac{\partial f}{\partial \overline{z}}\)? We have
\[
\begin{align} \frac{\partial f}{\partial z} & = \frac{\partial f}{\partial x} \frac{\partial x}{\partial {z}} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial {z}} = \left(u_x + iv_x\right)\frac{1}{2} + (u_y + iv_y)\cdot \left(\frac{1}{2i}\right) = \\ & = \frac{(u_x + v_y) + i(v_x - u_y)}{2} \\ \frac{\partial f}{\partial \overline{z}} & = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \overline{z}} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \overline{z}} = \left(u_x + iv_x\right) \frac{1}{2} + \left(u_y + iv_y\right)\cdot \left(-\frac{1}{2i}\right) = \\ & = \frac{1}{2} \left[\left(u_x - v_y\right) + i \left(v_x + u_y\right)\right] \end{align}
\]
Thus, the Cauchy-Riemann conditions in short form: \(\frac{\partial f}{\partial \overline{z}} = 0\). In this case \(df = \frac{\partial f}{\partial z} dz\), and moreover
\[
\frac{\partial f}{\partial z} = f'(z) = u_x - iu_y = v_y + iv_x
\]
For example, \(f(z) = z^2 = (x + iy)^2\). It is differentiable everywhere (this can be verified from the definition), and \(f(z) = x^2 - y^2 + 2ixy\), \(u = x^2 - y^2\), \(v = 2xy\). \(u_x = 2x\), \(u_y = -2y\), \(v_x = 2y\), \(v_y = 2x\).
\[
\frac{\partial f}{\partial z} = 2x + 2iy = 2z
\]
Write the Cauchy-Riemann conditions in polar coordinates.
Let \(z = re^{i\varphi }\), i.e. \(x = r\cos \varphi\), \(y = r\sin \varphi\). By the chain rule \[
\begin{align} u_r & = u_x \cos \varphi + u_y \sin \varphi , & u_\varphi & = -u_x r\sin \varphi + u_y r \cos \varphi , \\ v_r & = v_x \cos \varphi + v_y \sin \varphi , & v_\varphi & = -v_x r\sin \varphi + v_y r \cos \varphi . \end{align}
\] Substitute the Cauchy-Riemann conditions \(u_x = v_y\), \(u_y = -v_x\): \[
\begin{align} v_\varphi & = -v_x r \sin \varphi + v_y r\cos \varphi = u_y r \sin \varphi + u_x r \cos \varphi = r \cdot u_r, \\ u_\varphi & = -u_x r\sin \varphi + u_y r\cos \varphi = -v_y r \sin \varphi - v_x r \cos \varphi = -r \cdot v_r. \end{align}
\] In total, the Cauchy-Riemann conditions in polar coordinates: \[
u_r = \frac{1}{r} v_{\varphi }, \qquad v_r = -\frac{1}{r} u_{\varphi }
\]
\(u_r = \frac{1}{r} v_{\varphi }\), \(v_r = -\frac{1}{r} u_{\varphi }\).
Let \(D\) be a domain (i.e. an open connected set). A function \(f: D \to \mathbb {C}\) is called holomorphic (also called regular, analytic) if \(f\) is differentiable at every point of \(D\).
Let \(D \subset \mathbb {C}\) be a domain. Find all functions \(f\) holomorphic in \(D\) such that
\(v \equiv \operatorname {const}\) in \(D\);
\(\operatorname {arg}f(z) \equiv \operatorname {const}\) in \(D\).
\(v \equiv \operatorname {const}\), i.e. \(v_x = v_y = 0\) in \(D\). By the Cauchy-Riemann conditions \(u_x = v_y = 0\) and \(u_y = -v_x = 0\), i.e. \(u \equiv \operatorname {const}\) (the domain \(D\) is connected). In total \(f \equiv \operatorname {const}\).
\(\frac{v}{u} = \operatorname {tg}\operatorname {arg}f(z) = \alpha\) in \(D\), i.e. \(v = \alpha u\) (the case \(\operatorname {arg}f \equiv \pm \frac{\pi }{2}\), i.e. \(u \equiv 0\), is handled as in part 1). The Cauchy-Riemann conditions:
\[
\begin{cases} u_x = \alpha u_y \\ u_y = -\alpha u_x \end{cases}
\]
If there is at least one point in \(D\) where \(u_y \neq 0\), then \(-\alpha^2 = 1\), i.e. \(\alpha \in \left\{ i,-i\right\}\). Contradiction (since \(\alpha \in \mathbb {R}\)). I.e. \(u_x = u_y = v_x = v_y \equiv 0\), and again \(f \equiv \operatorname {const}\).
In both cases only \(f \equiv \operatorname {const}\).
Let \(v = x^{2}-y^{2}+2 x y\). Recover the holomorphic function \(f(z)\).
Let us check that \(v\) is harmonic (if this is not the case, the problem cannot be solved).
\[
\Delta v = v_{xx} + v_{yy} = 2 - 2 =0
\]
Indeed harmonic.
Further,
\[
\begin{align} u_y & = -v_x = -2x - 2y \; \Rightarrow \; u = -2xy - y^2 + \varphi (x) \\ -2y + \varphi '(x) = u_x & = v_y = -2y + 2x \; \Rightarrow \; \varphi (x) = x^2 + C \end{align}
\]
Let us obtain an expression for \(f\) in terms of \(z\). Note that \(z^2 = x^2 - y^2 + 2ixy\). We have
\[
f(z) = u + iv = -2xy - y^2 + x^2 + C + i\left(x^{2}-y^{2}+2 x y\right) = z^2 + iz^2 + C, \qquad C \in \mathbb {R}
\]
Find all points \(z = x+i y\) at which the following functions are differentiable:
\(|\overline{z}|^{2}\);
\(x^{2}-y^{2}-2 i x y\);
\(z \operatorname {Re} z\).
Find the domains in which the function \(f\) is holomorphic, if \[
f(z) = \left|x^{2}-y^{2}\right|+2 i|x y|, \quad z = x+i y
\]
Let \(f(z) = u(x, y)+i v(x, y)\), where \(z = x+i y\), and the real-valued functions \(u\) and \(v\) are differentiable at the point \(\left(x_{0}, y_{0}\right)\). Let \(z_{0} = x_{0}+i y_{0}\).
Find the set of limit points of the ratio \(\frac{f\left(z_{0}+\Delta z\right)-f\left(z_{0}\right)}{\Delta z}\) as \(\Delta z \rightarrow 0\).
Prove that if the limit
\[
\lim _{\Delta z \rightarrow 0} \operatorname {Re} \frac{f\left(z_{0}+\Delta z\right)-f\left(z_{0}\right)}{\Delta z},
\]
exists, then the function \(f\) is differentiable in the complex sense at the point \(z_{0}\).
Prove that if \(f(z)\) and \(\overline{f(z)}\) are both holomorphic in a domain \(D\), then \(f(z) \equiv \operatorname {const}\) in \(D\).
Prove that if the function \(f\) is defined in a domain \(D\), is \(n\) times differentiable in the complex sense, and \(f^{(n)}(z) \equiv 0\), then \(f\) is a polynomial of degree less than \(n\).